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| -rw-r--r-- | content/know/concept/cauchy-strain-tensor/index.pdc | 6 | ||||
| -rw-r--r-- | content/know/concept/cauchy-stress-tensor/index.pdc | 3 | ||||
| -rw-r--r-- | content/know/concept/hookes-law/index.pdc | 245 | ||||
| -rw-r--r-- | content/know/concept/navier-cauchy-equation/index.pdc | 115 |
4 files changed, 364 insertions, 5 deletions
diff --git a/content/know/concept/cauchy-strain-tensor/index.pdc b/content/know/concept/cauchy-strain-tensor/index.pdc index 2994674..f150723 100644 --- a/content/know/concept/cauchy-strain-tensor/index.pdc +++ b/content/know/concept/cauchy-strain-tensor/index.pdc @@ -54,7 +54,7 @@ $$\begin{aligned} Let us choose two nearby points in the deformed solid, and call them $\va{x}$ and $\va{x} + \va{a}$, where $\va{a}$ is a tiny vector pointing from one to the other. -Before the displacement, these points respectively had these positions, +Before the displacement, those points respectively had these positions, where we define $\va{A}$ as the "old" version of $\va{a}$: $$\begin{aligned} @@ -159,7 +159,7 @@ $$\begin{aligned} = 2 \va{a} \cdot \hat{u} \cdot \va{b} \end{aligned}$$ -The Cauchy strain tensor $\hat{u}$ is a second-order tensor, +The Cauchy strain tensor $\hat{u}$ is a second-rank tensor, and can alternatively be expressed like so: $$\begin{aligned} @@ -320,7 +320,7 @@ we remove it, and isolate the rest for $\delta(\dd{\va{S}})$: $$\begin{aligned} \boxed{ \delta(\dd{\va{S}}) - = \big( (\nabla \cdot \va{u}) \va{1} - \nabla \va{u} \big) \cdot \dd{\va{S}} + = \big( (\nabla \cdot \va{u}) \hat{1} - \nabla \va{u} \big) \cdot \dd{\va{S}} } \end{aligned}$$ diff --git a/content/know/concept/cauchy-stress-tensor/index.pdc b/content/know/concept/cauchy-stress-tensor/index.pdc index a26e2a8..080254d 100644 --- a/content/know/concept/cauchy-stress-tensor/index.pdc +++ b/content/know/concept/cauchy-stress-tensor/index.pdc @@ -108,7 +108,7 @@ $$\begin{aligned} } \end{aligned}$$ -The stress components $\sigma_{ij}$ can be written as a second-order tensor +The stress components $\sigma_{ij}$ can be written as a second-rank tensor (i.e. a matrix that transforms in a certain way), called the **Cauchy stress tensor** $\hat{\sigma}$: @@ -177,7 +177,6 @@ $$\begin{aligned} F_{s, i} = \oint_S \sum_j \sigma_{ij} \dd{S_j} = \int_V \sum_{j} \nabla_{\!j} \sigma_{ij} \dd{V} - = \int_V \nabla \cdot \vec{\sigma}_i \dd{V} \end{aligned}$$ In any case, the total force $\va{F}$ can then be expressed diff --git a/content/know/concept/hookes-law/index.pdc b/content/know/concept/hookes-law/index.pdc new file mode 100644 index 0000000..94ceb1b --- /dev/null +++ b/content/know/concept/hookes-law/index.pdc @@ -0,0 +1,245 @@ +--- +title: "Hooke's law" +firstLetter: "H" +publishDate: 2021-04-02 +categories: +- Physics +- Continuum physics + +date: 2021-03-09T17:31:37+01:00 +draft: false +markup: pandoc +--- + +# Hooke's law + +In its simplest form, **Hooke's law** dictates that +changing the length of an elastic object requires +a force that is proportional the desired length difference. +In its most general form, it gives a linear relationship +between the [Cauchy stress tensor](/know/concept/cauchy-stress-tensor/) $\hat{\sigma}$ +to the [Cauchy strain tensor](/know/concept/cauchy-strain-tensor/) $\hat{u}$. + +Importantly, all forms of Hooke's law are only valid for small deformations, +since the stress-strain relationship becomes nonlinear otherwise. + + +## Simple form + +The simple form of the law is traditionally quoted for springs, +since they have a spring constant $k$ giving the ratio +between the force $F$ and extension $x$: + +$$\begin{aligned} + \boxed{ + F + = k x + } +\end{aligned}$$ + +In general, all solids are elastic for small extensions, +and therefore also obey Hooke's law. +In light of this fact, we replace the traditional spring +with a rod of length $L$ and cross-section $A$. + +The constant $k$ depends on, among several things, +the spring's length $L$ and cross section $A$, +so for our generalization, we want a new parameter +to describe the proportionality independently of the rod's dimensions. +To achieve this, we realize that the force $F$ is spread across $A$, +and that the extension $x$ should be take relative to $L$. + +$$\begin{aligned} + \frac{F}{A} + = \Big( k \frac{L}{A} \Big) \frac{x}{L} +\end{aligned}$$ + +The force-per-area $F/A$ on a solid is the definition of **stress**, +and the relative elongation $x/L$ is the defintion of **strain**. +If $F$ acts along the $x$-axis, we can then write: + +$$\begin{aligned} + \boxed{ + \sigma_{xx} + = E u_{xx} + } +\end{aligned}$$ + +Where the proportionality constant $E$, +known as the **elastic modulus** or **Young's modulus**, +is the general material parameter that we wanted: + +$$\begin{aligned} + E + = k \frac{L}{A} +\end{aligned}$$ + +Due to the microscopic structure of some (usually crystalline) materials, +$E$ might be dependent on the direction of the force $F$. +For simplicity, we only consider **isotropic** materials, +which have the same properties measured from any direction. + +However, we are still missing something. +When a spring is pulled, +it becomes narrower as its coils move apart, +and this effect is also seen when stretching solids in general: +if we pull our rod along the $x$-axis, we expect it to deform in $y$ and $z$ as well. +This is described by **Poisson's ratio** $\nu$: + +$$\begin{aligned} + \boxed{ + \nu + \equiv - \frac{u_{yy}}{u_{xx}} + } +\end{aligned}$$ + +Note that $u_{yy} = u_{zz}$ because the material is assumed to be isotropic. +Intuitively, you may expect that the volume of the object is conserved, +but for most materials that is not accurate. + +In summary, for our example case with a force $F = T A$ pulling at the rod +along the $x$-axis, the full stress and strain tensors are given by: + +$$\begin{aligned} + \hat{\sigma} = + \begin{bmatrix} + T & 0 & 0 \\ + 0 & 0 & 0 \\ + 0 & 0 & 0 + \end{bmatrix} + \qquad + \hat{u} = + \begin{bmatrix} + T/E & 0 & 0 \\ + 0 & -\nu T/E & 0 \\ + 0 & 0 & -\nu T/E + \end{bmatrix} +\end{aligned}$$ + + +## General isotropic form + +The general form of Hooke's law is a linear relationship +between the stress and strain tensors: + +$$\begin{aligned} + \boxed{ + \hat{\sigma} + = 2 \mu \: \hat{u} + \lambda \Tr\!(\hat{u}) \: \hat{1} + } +\end{aligned}$$ + +Where $\Tr{}$ is the trace. +This is often written in index notation, +with the Kronecker delta $\delta_{ij}$: + +$$\begin{aligned} + \boxed{ + \sigma_{ij} + = 2 \mu u_{ij} + \lambda \delta_{ij} \sum_{k} u_{kk} + } +\end{aligned}$$ + +The constants $\mu$ and $\lambda$ are called the **Lamé coefficients**, +and are related to $E$ and $\nu$ in a way we can derive +by returning to the example with a tension $T = F/A$ along $x$. +For $\sigma_{xx}$, we have: + +$$\begin{aligned} + T + = \sigma_{xx} + &= 2 \mu u_{xx} + \lambda (u_{xx} + u_{yy} + u_{zz}) + \\ + &= \frac{2 \mu}{E} T + \frac{\lambda}{E} T - \frac{\nu \lambda}{E} (T + T) + \\ + &= \frac{T}{E} \Big( 2 \mu + \lambda (1 - 2 \nu) \Big) +\end{aligned}$$ + +Meanwhile, the other diagonal stresses $\sigma_{yy} = \sigma_{zz}$ +are expressed in terms of the strain like so: + +$$\begin{aligned} + 0 + = \sigma_{yy} + &= 2 \mu u_{yy} + \lambda (u_{xx} + u_{yy} + u_{zz}) + \\ + &= - \frac{2 \nu \mu}{E} T + \frac{\lambda}{E} T - \frac{\nu \lambda}{E} (T + T) + \\ + &= \frac{E}{T} \Big( \!-\! 2 \nu \mu + \lambda (1 - 2 \nu) \Big) +\end{aligned}$$ + +After dividing out superfluous factors from the two preceding equations, +we arrive at: + +$$\begin{aligned} + E + = 2 \mu + \lambda (1 - 2 \nu) + \qquad \quad + 2 \nu \mu + = \lambda (1 - 2 \nu) +\end{aligned}$$ + +Solving this system of equations for the Lamé coefficients +yields the following result: + +$$\begin{aligned} + \boxed{ + \lambda + = \frac{E \nu}{(1 - 2 \nu)(1 + \nu)} + \qquad \quad + \mu + = \frac{E}{2 (1 + \nu)} + } +\end{aligned}$$ + +Which can straightforwardly be inverted +to express $E$ and $\nu$ as a function of $\mu$ and $\lambda$: + +$$\begin{aligned} + \boxed{ + E + = \mu \frac{3 \lambda + 2 \mu}{\lambda + \mu} + \qquad \quad + \nu + = \frac{\lambda}{2 (\lambda + \mu)} + } +\end{aligned}$$ + +Hooke's law itself can also be inverted, +i.e. we can express the strain as a function of stress. +First, observe that the trace of the stress tensor satisfies: + +$$\begin{aligned} + \Tr\!(\hat{\sigma}) + = \sum_{i} \sigma_{ii} + = 2 \mu \sum_{i} u_{ii} + \lambda \sum_{i} \sum_{k} u_{kk} + = (2 \mu + 3 \lambda) \sum_{i} u_{ii} +\end{aligned}$$ + +Inserting this into Hooke's law +yields an equation that only contains one strain component $u_{ij}$: + +$$\begin{aligned} + \sigma_{ij} + = 2 \mu u_{ij} + \frac{\lambda}{2 \mu + 3 \lambda} \delta_{ij} \sum_{k} \sigma_{kk} +\end{aligned}$$ + +Which is therefore trivial to isolate for $u_{ij}$, +leading us to Hooke's inverted law: + +$$\begin{aligned} + \boxed{ + \begin{aligned} + u_{ij} + &= \frac{\sigma_{ij}}{2 \mu} - \frac{\lambda}{2 \mu (3 \lambda + 2 \mu)} \delta_{ij} \sum_{k} \sigma_{kk} + \\ + &= \frac{1 + \nu}{E} \sigma_{ij} - \frac{\nu}{E} \delta_{ij} \sum_{k} \sigma_{kk} + \end{aligned} + } +\end{aligned}$$ + + +## References +1. B. Lautrup, + *Physics of continuous matter: exotic and everyday phenomena in the macroscopic world*, 2nd edition, + CRC Press. diff --git a/content/know/concept/navier-cauchy-equation/index.pdc b/content/know/concept/navier-cauchy-equation/index.pdc new file mode 100644 index 0000000..d3802d3 --- /dev/null +++ b/content/know/concept/navier-cauchy-equation/index.pdc @@ -0,0 +1,115 @@ +--- +title: "Navier-Cauchy equation" +firstLetter: "N" +publishDate: 2021-04-02 +categories: +- Physics +- Continuum physics + +date: 2021-04-02T15:04:55+02:00 +draft: false +markup: pandoc +--- + +# Navier-Cauchy equation + +The **Navier-Cauchy equation** describes **elastodynamics**: +the movements inside an elastic solid +in response to external forces and/or internal stresses. + +For a particle of the solid, whose position is given by the displacement field $\va{u}$, +Newton's second law is as follows, +where $\dd{m}$ and $\dd{V}$ are the particle's mass and volume, respectively: + +$$\begin{aligned} + \va{f^*} \dd{V} + = \pdv[2]{\va{u}}{t} \dd{m} + = \rho \pdv[2]{\va{u}}{t} \dd{V} +\end{aligned}$$ + +Where $\rho$ is the mass density, +and $\va{f^*}$ is the effective force density, +defined from the [Cauchy stress tensor](/know/concept/cauchy-stress-tensor/) $\hat{\sigma}$ +like so, with $\va{f}$ being an external body force, e.g. from gravity: + +$$\begin{aligned} + \va{f^*} + = \va{f} + \nabla \cdot \hat{\sigma}^\top +\end{aligned}$$ + +We can therefore write Newton's second law as follows, +while switching to index notation, +where $\nabla_j = \pdv*{x_j}$ is the partial derivative +with respect to the $j$th coordinate: + +$$\begin{aligned} + f_i + \sum_{j} \nabla_j \sigma_{ij} + = \rho \pdv[2]{u_i}{t} +\end{aligned}$$ + +The components $\sigma_{ij}$ of the Cauchy stress tensor +are given by [Hooke's law](/know/concept/hookes-law/), +where $\mu$ and $\lambda$ are the Lamé coefficients, +which describe the material: + +$$\begin{aligned} + \sigma_{ij} + = 2 \mu u_{ij} + \lambda \delta_{ij} \sum_{k} u_{kk} +\end{aligned}$$ + +In turn, the components $u_{ij}$ of the +[Cauchy strain tensor](/know/concept/cauchy-strain-tensor/) +are defined as follows, +where $u_i$ are once again the components of the displacement vector $\va{u}$: + +$$\begin{aligned} + u_{ij} + = \frac{1}{2} \big( \nabla_i u_j + \nabla_j u_i \big) +\end{aligned}$$ + +To derive the Navier-Cauchy equation, +we start by inserting Hooke's law into Newton's law: + +$$\begin{aligned} + \rho \pdv[2]{u_i}{t} + %= f_i + \sum_{j} \nabla_j \sigma_{ij} + &= f_i + 2 \mu \sum_{j} \nabla_j u_{ij} + \lambda \sum_{j} \nabla_j \bigg( \delta_{ij} \sum_{k} u_{kk} \bigg) + \\ + &= f_i + 2 \mu \sum_{j} \nabla_j u_{ij} + \lambda \nabla_i \sum_{j} u_{jj} +\end{aligned}$$ + +And then into this we insert the definition of the strain components $u_{ij}$, yielding: + +$$\begin{aligned} + \rho \pdv[2]{u_i}{t} + &= f_i + \mu \sum_{j} \nabla_j \big( \nabla_i u_j + \nabla_j u_i \big) + \lambda \nabla_i \sum_{j} \nabla_j u_{j} +\end{aligned}$$ + +Rearranging this a bit leads us to the Navier-Cauchy equation written in index notation: + +$$\begin{aligned} + \boxed{ + \rho \pdv[2]{u_i}{t} + = f_i + \mu \sum_{j} \nabla_j^2 u_i + (\mu + \lambda) \nabla_i \sum_{j} \nabla_j u_j + } +\end{aligned}$$ + +Traditionally, it is written in vector notation instead, +in which case it looks like this: + +$$\begin{aligned} + \boxed{ + \rho \pdv[2]{\va{u}}{t} + = \va{f} + \mu \nabla^2 \va{u} + (\mu + \lambda) \nabla (\nabla \cdot \va{u}) + } +\end{aligned}$$ + +A special case is the **Navier-Cauchy equilibrium equation**, +where the left-hand side is just zero. +That version describes **elastostatics**: the deformation of a solid at rest. + + +## References +1. B. Lautrup, + *Physics of continuous matter: exotic and everyday phenomena in the macroscopic world*, 2nd edition, + CRC Press. |